# A 3-minute algebra problem

I have just taken advance math test level 2 and there are several problems that have been bugging me. This is the first question:

If $x,y>0$, then determine the value of $x$ that satisfies the system of equations: \begin{align} x^2+y^2-xy&=3\\ x^2-y^2+\sqrt{6}y&=3\\ \end{align}

I can answer this problem using a 'standard' algebra but it takes time more than 3 minutes. For the sake of curiosity, is there a way to answer this problem less than 3 minutes?

• (I) - (II) gives you $2y = x + \sqrt 6$ after dividing by $y$. – G. Bach Apr 27 '14 at 16:17
• @G.Bach I did notice that and that's also my way to answer this problem but it took time more than 3 minutes – Anastasiya-Romanova 秀 Apr 27 '14 at 16:19

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$

Substracting second equation to the first you obtain $$2y^2-xy-\sqrt{6}y=0$$ Then, factoring $$y(2y-x-\sqrt{6})=0$$ Thus $2y-x-\sqrt{6}=0$ cause $x,y$ are greater than $0$. So $y=\frac{x+\sqrt{6}}{2}$, putting this expression into the second equation you have \begin{align} x^2-\left(\frac{x+\sqrt{6}}{2}\right)^2+\sqrt{6}\left(\frac{x+\sqrt{6}}{2}\right)&=3 \\ x^2-\frac{x^2+2\sqrt{6}x+6}{4}+\frac{\sqrt{6}}{2}x+3&=3\\ \frac{3}{4}x^2-\frac{3}{2}&=0\\ \end{align} Whose positive root is $x=\sqrt{2}$.

Take G.Bach's hint so that $2y=x+\sqrt 6$

Now add the two equations to obtain $$2x^2+(\sqrt 6-x)y=6$$ Substitute for $y$: $$2x^2+\frac 12(\sqrt 6-x)(\sqrt 6+x)=6=\frac 32 x^2+3$$

So that $x^2=2$

• The essential thing to simplify the arithmetic was to get rid of the term in $y^2$ before substituting. The fact that this gave a simple solution (rather than a more complex quadratic) was a bonus. – Mark Bennet Apr 27 '14 at 16:37